Final answer:
The ratio of the probability of losing to the probability of winning in the play of 2 dice is 2/5.
Step-by-step explanation:
To find the ratio of the probability of losing to the probability of winning, we need to calculate the probability of each outcome. Let's start with the probability of losing. The thrower loses if the first throw is a total of 2, 4, or 12. Since there are only two dice, there are only 36 possible outcomes. The combinations that result in a total of 2 are (1,1), (1, 1), (2,1), and (1, 2), which is 4 out of 36, or 1/9. Similarly, the combinations for a total of 4 and 12 are (1,3), (1,3), (1,2), (2,1), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3), (6,6), (6,6), (5,5), and (5,5), which is also 1/9 each. Therefore, the probability of losing is 1/9 + 1/9 + 1/9 = 3/9 = 1/3.
Now let's calculate the probability of winning. The thrower wins if the first throw is a total of 5 or 11. The combination for a total of 5 is (1,4), (4,1), (2,3), and (3,2), which is 4 out of 36, or 1/9. The combinations for a total of 11 are (5,6) and (6,5), which is 2 out of 36, or 1/18 each. Therefore, the probability of winning is 1/9 + 1/18 + 1/18 = 5/18.
Now we can calculate the ratio of the probability of losing to the probability of winning. The ratio is obtained by dividing the probability of losing (1/3) by the probability of winning (5/18):
Ratio = (1/3) / (5/18) = (1/3) * (18/5) = 6/15 = 2/5